Problem: Simplify and expand the following expression: $ \dfrac{r + 6}{r - 2}-\dfrac{2r - 7}{2r - 3} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(r - 2)(2r - 3)$ Multiply the first term by $\dfrac{2r - 3}{2r - 3}$ $ \begin{align*} \dfrac{r + 6}{r - 2} \times \dfrac{2r - 3}{2r - 3} & = \dfrac{(r + 6)(2r - 3)}{(r - 2)(2r - 3)} \\ & = \dfrac{2r^2 + 9r - 18}{(r - 2)(2r - 3)}\end{align*} $ Multiply the second term by $\dfrac{r - 2}{r - 2}$ $ \begin{align*} \dfrac{2r - 7}{2r - 3} \times \dfrac{r - 2}{r - 2} & = \dfrac{(2r - 7)(r - 2)}{(2r - 3)(r - 2)} \\ & = \dfrac{2r^2 - 11r + 14}{(2r - 3)(r - 2)}\end{align*} $ Now we have: $ = \dfrac{2r^2 + 9r - 18}{(r - 2)(2r - 3)} - \dfrac{2r^2 - 11r + 14}{(2r - 3)(r - 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2r^2 + 9r - 18 - (2r^2 - 11r + 14)}{(r - 2)(2r - 3)} $ $ = \dfrac{2r^2 + 9r - 18 - 2r^2 + 11r - 14}{(r - 2)(2r - 3)} $ $ = \dfrac{20r - 32}{(r - 2)(2r - 3)}$ Expand the denominator: $ = \dfrac{20r - 32}{2r^2 - 7r + 6}$